Ctg²x+2√3ctgx+3sin²x=-3sin²(x-3/2π)
sin²(x-3/2π)=(sin(x-3/2π))²=(sin(-(3/2π-x)))²=(-sin(3/2π-x))²=(cosx)²=cos²x
ctg²x+2√3ctgx+3sin²x=-3cos²x
ctg²x+2√3ctgx+3sin²x+3cos²x=0
ctgx=t
t²+2√3t+1=0
t=-√3
ctgx=-√3
x=π-arcctg√3+πn, n∈Z
x=π-π/6+πn, n∈Z
x=5π/6+πn, n∈Z
-11π/2≤5π/5+πn≤-4π
-11π/2-5π/6≤πn≤-4π-5π/6
-38π/6≤πn≤-29π/6
-38/6≤n≤-29/6
n=-5, n=-6
x₁=5π/6-5π, x₁=-25π/6
x₂=5π/6-6π, x₂=-31π/6